algebraically solve 3cosx +sinx =2 for O<x<360

Question

I am trying to solve this equation algebraically so far I have

3cosx +sinx =2

3cosx            =2 -sinx

9cos^2x         =(2-sinx)^2

Answer 1

9cos²(x)=4-4sin(x)+sin²(x),

10sin²(x)-4sin(x)=5,

sin²(x)-0.4sin(x)=0.5,

sin²(x)-0.4sin(x)+0.04=0.54,

(sin(x)-0.2)²=0.54,

sin(x)-0.2=±√0.54,

sin(x)=0.2±√0.54, so sin(x)=0.9348 or -0.5348, making x=69.2034° or -32.3335°.

We need angles between 0° and 360°, so we could have 4 solutions:

69.2034°, 180°-69.2034°=110.7966°, 360°-32.3335°=327.6665°, 180°+32.3335°=212.3335°.

Now we need to test each by substituting for x in the original equations. This test eliminates 212.3335° and 110.7966°.

So there are only two valid solutions in the given range: 69.2034°, 327.6665°. The other proposed solutions were created as the result of squaring in the method.